The given heat is the transfer of energy between two bodies at different temperatures. The one with a higher temperature gives up heat to the one with a lower temperature. Whether a body yields or absorbs heat, its temperature or its physical state can vary depending on the mass and characteristics of the material from which it is made..
A good example is in a steaming cup of coffee. The metal spoon with which the sugar is stirred heats up. If it is left in the cup long enough, coffee and a metal spoon will end up equalizing their temperatures: the coffee will have cooled down and heat will have been given to the spoon. Some heat will have passed into the environment, since the system is not insulated.
As the temperatures become equal, the thermal equilibrium.
If you did the same test with a plastic teaspoon, you would surely notice that it does not heat up as quickly as the metal one, but it will eventually also come into balance with the coffee and everything around it..
This is because metal conducts heat better than plastic. On the other hand, surely coffee yields heat at a different rate than hot chocolate or other beverage. So the heat given or absorbed by each object depends on what material or substance it is made of..
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Heat always refers to the flow or transit of energy between one object and another, due to the difference in temperature.
That is why we speak of transferred heat or absorbed heat, since by adding or extracting heat or energy in some way, it is possible to modify the temperature of an element.
The amount of heat that the hottest object gives off is usually called Q. This value is proportional to the mass of said object. A body with a large mass is capable of yielding more heat than another with a lower mass..
Another important factor in calculating the heat released is the difference in temperature experienced by the object that releases the heat. It is denoted as ΔT and it is calculated like this:
ΔT = TF - Tor
Finally, the amount of heat transferred also depends on the nature and characteristics of the object, which are quantitatively summarized in a constant called specific heat of the material, denoted as c.
So finally the expression for the transferred heat is the following:
Qyielded = - m.c.ΔT
Giving in is symbolized by a negative sign.
Specific heat is the amount of heat necessary to raise the temperature of 1 g of substance by 1 ºC. It is an intrinsic property of the material. Its units in the International System are: Joule / kg. K (Joule between kilogram x temperature in degrees Kelvin).
Heat capacity C is a linked concept, but slightly different, since the mass of the object is involved. Heat capacity is defined as follows:
C = mc
Its units in S.I. they are Joule / K. So the heat released can also be expressed equivalently as:
Q = -C. ΔT
To calculate the heat given off by an object, it is necessary to know the following:
- The specific heat of the substance that gives up heat.
- The mass of said substance
- The final temperature to obtain
Specific heat values for many materials have been determined experimentally and are available in tables.
Now, if this value is not known, it is possible to obtain it with the help of a thermometer and water in a thermally insulated container: the calorimeter. A diagram of this device is shown in the figure that accompanies exercise 1.
A sample of the substance at a certain temperature is immersed in a previously measured quantity of water. The final temperature is measured and the specific heat of the material is determined with the values obtained..
By comparing the result with the tabulated values, it can be known which substance it is. This procedure is called calorimetry.
The heat balance is carried out by conserving energy:
Q yielded + Q absorbed = 0
A 0.35 kg piece of copper is introduced at a temperature of 150ºC in 500 mL of water at a temperature of 25ºC. Find:
a) The final equilibrium temperature
b) How much heat flows in this process?
ccopper = 385 J / kg. ºC
cwater = 4180 J / kg. ºC
Water density: 1000 kg / m3
a) Copper gives up heat while water absorbs it. As the system is considered closed, only the water and the sample intervene in the heat balance:
Q yielded = Q absorbed
On the other hand, it is required to calculate the mass of 500 mL of water:
500 mL = 0.5 L = 0.0005 m3
With these data the mass of the water is calculated:
mass = density x volume = 1000 kg / m3 . 0.0005 m3 = 0.5 kg
The equation for the heat in each substance is raised:
Qyielded = -mcopper . ccopper. ΔT = -0.35 kg. 385 J / kg. ° C . (TF -150 ºC) = -134.75 (TF - 150) J
Qabsorbed = mWater . cWater. ΔT = 0.5 kg. 4186 J / kg. ºC. (TF -25 ºC) = 2093 (TF -25) J
Equating the results, we have:
2093 (TF - 25) = -134.75 (TF - 150)
It is a linear equation with one unknown, whose solution is:
TF = 32.56 ºC
b) The amount of heat that flows is the heat transferred or the heat absorbed:
Q yielded = - 134.75 (32.56 - 150) J = 15823 J
Q absorbed = 2093 (32.56 - 25) J = 15823 J
A 100 g piece of copper is heated in an oven at a temperature Tor and then it is introduced into a 150 g copper calorimeter containing 200 g of water at 16ºC. The final temperature once in equilibrium is 38ºC. When the calorimeter and its contents are weighed, it is found that they have evaporated 1.2 g of water What was the initial temperature Tor?
Data: the latent heat of vaporization of water is Lv = 2257 kJ / kg
This exercise differs from the previous one, since it must be considered that the calorimeter also absorbs heat. The heat released by the piece of copper is invested in all of the following:
- Heat the water in the calorimeter (200 g)
- Heat the copper from which the calorimeter is made (150 g)
- Evaporate 1.2 grams of water (energy is also needed for a phase change).
Qyielded = -100 x 1 x 10 -3 kg. 385 J / kg. ºC. (38 - Tor ) ºC = -38.5. (38 - Tor) J
Q absorbed by calorimeter = Q absorbed by water + Q vaporization + Q absorbed by copper
0.2 kg .4186 J / kg ºC. (38 - 16 ºC) + 1.2 x 10-3 kg. 2257000 J / kg +0.150 kg .385 J / kg .ºC. (38 - 16 ºC) =
18418.4 +2708.4 + 1270.5 J = 22397.3 J
Therefore:
-38.5. (38 - Tor) = 22397.3
Tor = 619.7 ºC
The heat needed to bring the 1.2 g of water up to 100ºC could also have been considered, but it is quite a small amount in comparison..
Very good work!